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Question Number 139918 by mathsuji last updated on 02/May/21
1+cos2°+cos4°+cos6°+...+cos88°
Answered by mr W last updated on 02/May/21
∑44k=0(cos2kπ180+isin2kπ180)=∑44k=0e2kπ180i=1−e2πi180×451−e2πi180=1−i(1−cosπ90)−isinπ90=(1−i)[(1−cosπ90)+isinπ90](1−cosπ90)2+sin2π90=1−cosπ90+sinπ90−i(1−cosπ90+sinπ90)(1−cosπ90)2+sin2π90=1−cosπ90+sinπ90−i(1−cosπ90+sinπ90)2(1−cosπ90)1+cos2°+cos4°+...+cos88°==∑44k=0cos2kπ180=1−cosπ90+sinπ902(1−cosπ90)=12(1+cotπ180)=12(1+cot1°)
Commented by mathsuji last updated on 02/May/21
thankyouverymuchSir..
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